Problem: In right triangle $PQR$, we have $\angle Q = \angle R$ and $PR = 6\sqrt{2}$.  What is the area of $\triangle PQR$?
Answer: A triangle can't have two right angles, so a right triangle with two congruent angles must have congruent acute angles.  That is, $\triangle PQR$ must be an isosceles right triangle with acute angles at $Q$ and $R$.  Therefore, $PQ=PR=6\sqrt{2}$,  and $[QRP]=(QP)(RP)/2 = (6\sqrt{2})(6\sqrt{2})/2 = (6\cdot 6\cdot\sqrt{2}\cdot \sqrt{2})/2 =\boxed{36}$.

[asy]

unitsize(1inch);

pair P,Q,R;

P = (0,0);

Q= (1,0);

R = (0,1);

draw (P--Q--R--P,linewidth(0.9));

draw(rightanglemark(Q,P,R,3));

label("$P$",P,S);

label("$Q$",Q,S);

label("$R$",R,N);

[/asy]